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Forces in 10 Minutes

Review forces in 10 minutes or less + 60 questions to help you master forces.
Contents

In this section, we will master Newton’s Second Law (Fnet = ma). We will also cover common areas of struggle, such as drawing FBDs or finding net forces.

Concept overview

  1. Newtons law
  2. Drawing FBD’s properly
  3. Resolving vectors
  4. Finding net forces correctly
  5. Solving force problems

(1) Newton’s Laws

(1.1) Law 1: Inertia.

An object in motion will stay in motion. An object at rest will stay at rest.

Alternate explanation: all objects want to resist changes in their motion. The more inertia (mass) an object has, the more it can resist change. Think of how hard it is to bring a plane to a complete stop.

(1.2) Law 2: Fnet = ma.

Any horizontal or vertical NET force acting on a mass will cause an acceleration.

Net vertical force is the sum of ALL forces vertically. The net horizontal force is the sum of ALL forces horizontally.

(1.3) Law 3: Equal and opposite forces.

Push down on your desk. Does it move? It doesn’t, because it’s pushing back up at you with the same force. When two objects interact with each other, they both apply the same force to each other.

For every action, there is an equal and opposite reaction.

Alternate explanation: think of a truck hitting a motorcycle. They will experience the same force, but since the motorcycle has less mass it will experience greater acceleration.

(2) FBD – Force Body Diagrams

If you want to solve questions correctly, your best bet is to start by drawing an FBD. Here are the rules for drawing common forces.

  1. Weight ALWAYS points straight down. Draw the vector from the center of the object.
  2. The normal force is ALWAYS 90° to the surface. Draw the vector from the bottom of the object.
  3. Friction always points opposite to the direction of motion (except in circular motion, in which it points inwards)
  4. Vectors can be at angles, but don’t draw the vector components unless asked to.
  5. Vecotrs of equal magnitude should be of equal length.

(2.1) FBD’s of Common Situations

  • Mass in free fall, neglecting air resistance
  • Mass falling at terminal velocity
  • Box sliding horizontally with friction
  • Box sliding down a ramp with friction
  • Elevator going up and slowing down
  • Two different masses on opposite sides of a pulley

(3) Mass vs. Weight

Mass is a scalar quantity. It remains constant EVERYWHERE.

Weight is a vector quantity, that describes the force on the mass due to gravity. Weight changes with gravity. Gravity varies from planet to planet.

  • Weight = Fw = mg

(4) Choosing Directions

You might have seen gravity to be positive or negative. The direction of positive and negative motion is arbitrarily chosen. That means you can choose if up means +/- or if left means +/-.

But you need to be consistent. Indicate the directions on your FBD. If you set downwards to be positive, then gravity will be positive. Thus if the object is moving up, then velocity is negative.

(5) Resolving Vectors

We only use horizontal and vertical forces when solving force problems. Therefore, if you are using a vector that is at an angle, you need to break (resolve) it into x and y components.

Use this trig shortcut: Given the vector A, look at the component of the triangle you are trying to find.

  • If that component is opposite to the angle –> use Asinø
  • If that component is adjacent to the angle –> use Acosø

(6) Net Forces vs. Equilibrium

The net force is the SUM of ALL forces in either the X or Y direction.

A object in Equilibrium is when there is 0 Net force. This means all the forces cancel out in that particular direction.

Since Fnet = 0 = ma –> this implies the net acceleration of the object is 0. This either means the object is not moving OR moving at constant velocity. For example, a book laying on the table has 0 net force, because the weight vector and normal vector cancel out.

–> 0 Net Force = 0 acceleration = constant velocity

(7) Equations of Common Forces

(A) Friction = ƒ = Fƒ = µkN or µsN

  • N = normal force
  • µk = coefficient of kinetic friction (for when the object is moving)
  • µs = coefficient of static friction (for when the object is not moving
  • µs > µk
  • PRO TIP: on a ramp –> µs = tanø

(B) Hooke’s Law = Spring Force = Fs = kx

  • k = spring constant
  • x = compression or stretch of the spring

(C) Centeripetal force = Fc = mac = mv2/r

  • v = tangential velocity
  • r = radius of the circle

(D) Gravitational Force between two masses = Fg = (Gm1m2)/r2

  • G = gravitational constant ≈ 6.67 × 10-11
  • m1 = mass of object 1
  • m2 = mass of object 2
  • r = distance between m1 and m2

(8) Solving Force Problems (Tips and Tricks)

Students struggle the most on this. Using the step below you can solve ALL force problems, it just comes down to practice. Simply put, to solve the force problem:

  1. Turn the word problem into a FBD.
  2. Use the FBD to find either the horizontal or vertical net force.
  3. Lastly set the net force equal to ma, to solve for the unknown variable.

There are 100’s of types of problems involving forces. There are 60 practice questions, based on the actual AP Physics 1 exam, below to help you tackle the most common forces questions. These will likely show up on your test.

(8.1) Common Force Questions

Listed below is a list of common force problems, you should know how to solve all of these. Use the 3 step method above to solve.

  • Box being pushed across a floor (with friction)
  • Box being pull by cord at an angle above the horizontal
  • Box sliding down at incline (with friction)
  • Pulley systems (Atwood’s machine)
  • Pulley system on table with one mass hanging off
  • Elevators accelerating up/down
  • Cars going around in a circle with friction
  • Cars going on banked curves without friction (finding max speed)

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(9) 10 practice questions to help you master forces

Solve all 10 questions first. Answers are given at the end of the section.

(1) Find the downwards acceleration of an elevator, given that the ratio of a person’s stationary weight to their weight in the elevator is 5:4.

⇨ concepts involved: gravity, net forces, elevators
⇨ difficulty: easy

(2) Determine the force needed to push a 150kg body up a smooth 30° incline with an acceleration of 6 m/s2.

⇨ concepts involved: inclines, net forces
⇨ difficulty: easy

(3) A 1kg and unknown mass (M) hangs on opposite sides of the pulley suspended from the ceiling. When the masses are released, M accelerates down at 5 m/s2. What is M?

⇨ concepts involved: pulley systems, multiple mass systems
⇨ difficulty: medium

(4) A sled moves with constant speed down a sloped hill. The angle of the hill with respect to the horizontal is 10.0°. What is the coefficient of kinetic friction between the sled and the hill’s surface?

⇨ concepts involved: inclines, friction, net forces
⇨ difficulty: medium

(5) Conceptual problem: A ladder is leaning against a wall at an angle. Which of the following forces must have the same magnitude as the frictional force exerted on the ladder by the floor?

(a) The force of gravity on the ladder
(b) The normal force exerted on the ladder by the floor
(c) The frictional force exerted on the ladder by the wall
(d) The normal force exerted on the ladder by the wall

⇨ concepts involved: Newton’s laws
⇨ difficulty: medium

(6) 3 blocks of masses 5, 4, and 3kg are placed side by side in that order. A 25N force applied on the 5kg block accelerates all 3 blocks together to the right. (a) Find the acceleration of the blocks. (b) Find the normal force the 4kg block exerts on the 3kg block.

⇨ concepts involved: multi body systems, acceleration, normal forces
⇨ difficulty: medium

(7) Conceptual Problem: A truck of mass 3500kg hits the back of a small car of mass 1400kg. Which car exerted more force on the other and why?

(a) The truck exerts more force on the car because it has more mass.
(b) The tuck exerts more force on the car because it has more acceleration.
(c) The car exerts more force on the truck because it has less mass.
(d) The car exerts more force on the truck because it has more acceleration
(e) They exert the same force, and the resulting acceleration of the car is greater.

⇨ concepts involved: Newton’s laws
⇨ difficulty: Hard

(8) A block of mass m is accelerated across a rough surface by a force of magnitude F exerted at an angle θ above the horizontal. The frictional force between the block and surface is ƒ. Find the acceleration of the block (as an equation).

⇨ concepts involved: FBD, F = ma, inclines
⇨ difficulty: medium

(9) Conceptual Problem: A clothesline is stretched between two trees. A tire hangs in the middle of the line, and the two halves of the line make equal angles with the horizontal. The tension in the line is

(a) half the tire’s weight
(b) is equal to the tire’s weight divided by 9.18 m/s2
(c) is less than half the tire’s weight
(d) is equal to the tire’s weight
(e) is more than half the tire’s weight.

⇨ concepts involved: tension
⇨ difficulty: medium

(10) A child on Earth has a weight of 500N. Determine the weight of the child if the earth was to triple in both mass and radius (3M and 3r).

⇨ concepts involved: Gravitational forces, weight
⇨ difficulty: medium

Answers:

  1. 2 m/s2
  2. 1650 N
  3. 3 kg
  4. tan(10) or .176
  5. C.
  6. 2.1 m/s2 and 6.25 N
  7. E.
  8. a = (Fcosθ -ƒ)/m
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Jason Kuma
Jason Kuma

Founder, Writer, Physic B.S, Business B.A USC, Fremont CA

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